Integral de $$$2 t e^{- 5 t}$$$

Encontre $$$\int 2 t e^{- 5 t}\, dt$$$.

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=2$$$ e $$$f{\left(t \right)} = t e^{- 5 t}$$$:

$${\color{red}{\int{2 t e^{- 5 t} d t}}} = {\color{red}{\left(2 \int{t e^{- 5 t} d t}\right)}}$$

Para a integral $$$\int{t e^{- 5 t} d t}$$$, use integração por partes $$$\int \operatorname{u} \operatorname{dv} = \operatorname{u}\operatorname{v} - \int \operatorname{v} \operatorname{du}$$$.

Sejam $$$\operatorname{u}=t$$$ e $$$\operatorname{dv}=e^{- 5 t} dt$$$.

Então $$$\operatorname{du}=\left(t\right)^{\prime }dt=1 dt$$$ (os passos podem ser vistos ») e $$$\operatorname{v}=\int{e^{- 5 t} d t}=- \frac{e^{- 5 t}}{5}$$$ (os passos podem ser vistos »).

A integral torna-se

$$2 {\color{red}{\int{t e^{- 5 t} d t}}}=2 {\color{red}{\left(t \cdot \left(- \frac{e^{- 5 t}}{5}\right)-\int{\left(- \frac{e^{- 5 t}}{5}\right) \cdot 1 d t}\right)}}=2 {\color{red}{\left(- \frac{t e^{- 5 t}}{5} - \int{\left(- \frac{e^{- 5 t}}{5}\right)d t}\right)}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(t \right)}\, dt = c \int f{\left(t \right)}\, dt$$$ usando $$$c=- \frac{1}{5}$$$ e $$$f{\left(t \right)} = e^{- 5 t}$$$:

$$- \frac{2 t e^{- 5 t}}{5} - 2 {\color{red}{\int{\left(- \frac{e^{- 5 t}}{5}\right)d t}}} = - \frac{2 t e^{- 5 t}}{5} - 2 {\color{red}{\left(- \frac{\int{e^{- 5 t} d t}}{5}\right)}}$$

Seja $$$u=- 5 t$$$.

Então $$$du=\left(- 5 t\right)^{\prime }dt = - 5 dt$$$ (veja os passos »), e obtemos $$$dt = - \frac{du}{5}$$$.

A integral pode ser reescrita como

$$- \frac{2 t e^{- 5 t}}{5} + \frac{2 {\color{red}{\int{e^{- 5 t} d t}}}}{5} = - \frac{2 t e^{- 5 t}}{5} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{5}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=- \frac{1}{5}$$$ e $$$f{\left(u \right)} = e^{u}$$$:

$$- \frac{2 t e^{- 5 t}}{5} + \frac{2 {\color{red}{\int{\left(- \frac{e^{u}}{5}\right)d u}}}}{5} = - \frac{2 t e^{- 5 t}}{5} + \frac{2 {\color{red}{\left(- \frac{\int{e^{u} d u}}{5}\right)}}}{5}$$

A integral da função exponencial é $$$\int{e^{u} d u} = e^{u}$$$:

$$- \frac{2 t e^{- 5 t}}{5} - \frac{2 {\color{red}{\int{e^{u} d u}}}}{25} = - \frac{2 t e^{- 5 t}}{5} - \frac{2 {\color{red}{e^{u}}}}{25}$$

Recorde que $$$u=- 5 t$$$:

$$- \frac{2 t e^{- 5 t}}{5} - \frac{2 e^{{\color{red}{u}}}}{25} = - \frac{2 t e^{- 5 t}}{5} - \frac{2 e^{{\color{red}{\left(- 5 t\right)}}}}{25}$$

Portanto,

$$\int{2 t e^{- 5 t} d t} = - \frac{2 t e^{- 5 t}}{5} - \frac{2 e^{- 5 t}}{25}$$

Simplifique:

$$\int{2 t e^{- 5 t} d t} = \frac{2 \left(- 5 t - 1\right) e^{- 5 t}}{25}$$

Adicione a constante de integração:

$$\int{2 t e^{- 5 t} d t} = \frac{2 \left(- 5 t - 1\right) e^{- 5 t}}{25}+C$$

$$$\int 2 t e^{- 5 t}\, dt = \frac{2 \left(- 5 t - 1\right) e^{- 5 t}}{25} + C$$$A